For electron moving with speed of 20 m/s and accelerated by 120 V p.D. de-Broglie wavelength can be easily obtained.

Explanation

Let stationary electron having mass m and charge e is accelerated by voltage V.
Kinetic energy of electron =voltage (p.d.)
K=eV
But `K=(1)/(2)mv^(2)`
Multiplying and dividing by m,
`=(1)/(2)(m^(2)v^(2))/(m)`
p=mV
`p^(2)=m^(2)V^(2)`
`therefore K=(p^(2))/(2m)`
`therefore p^(2)=2mK`
`therefore p=sqrt(2mK)`
k=eV
`therefore p=sqrt(2meV)`
`therefore lambda=(h)/(p)=(h)/(sqrt(2meV))`
`therefore lambda=(h)/(sqrt(2meV))`
Substituting value
`m=9.11xx10^(-31)`,e`=1.6xx10^(-19)`, V=120 V,
`h=6.625xx10^(-34)`
`lambda=(6.625xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx120))`
0.112 nm